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anonymous
 5 years ago
Hello!! Is anyone good with evaluating indefinite integrals?
anonymous
 5 years ago
Hello!! Is anyone good with evaluating indefinite integrals?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This one seems easy but I could sure use confirmation on the correct answer!! Lol.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0uh uhs, amistre is here :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I couldn't figure out how to post a division with out the sign!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2give me a T; give me an R; give me an IG...whats that spell? trouble?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lols nah, spells awesomeness

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2post it your best and well decipher it :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0^ in the equation editor and like amistre said it's not a big deal

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lols all yours amistre

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2integrate arcsin + C? or is that your answer that yougot?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That's my answer  I just have to evaluate the integral

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2whats the integral so that we can pervue its intricate innards :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{\sqrt{1  x^2}}\] is this the integral?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0without the integral sign...or the dx...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\frac\[\int\limits_{?}^{?}\frac{1}{1+x^2} dx\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2i saw it in a movie lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lols "down the rabbit hole"?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2:) tan^1(x) + C...are there any bounds to evaluate?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[[\int\limits] dx 1 + x2 = \arctan x + C \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Was that all I needed to refer to?! I have it in my notes but the 'dx' was in the num

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2its like learning greek all over again ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0something like this? \[\int\limits_{}^{}\frac{1}{x^2 + 1}dx\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2\[\int\limits_{} \frac{1}{1+x^2} dx \rightarrow \tan^{1}(x) + C\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for answering this question, amistre...i give you...a MEDAL

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2yay!! ..... its prolly foiled covered chocolate...so I accept :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lols how'd you guess? enjoy! cheers

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Will arctan x + c be correct?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2as long as you havent left anyting out, yes; but arctan is a dated term

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2i would know; im dated ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can you guys help me with another problem/confirm my answer being right or wrong?!!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2perhaps, but display the question first; just makes it easier :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Eval the Indefinite Integral :\[\int\limits\limits_{?}^{?}\frac{x^3 dx}{\sqrt[4]{5x^41}}\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2you want a 15 for that :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got: 1/15 (5x^4)^3/4 + C

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2its refered to as "u substitution" u = 5x^4  1; du = 20x^3 dx dx = du/20x^3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{15} (5x^{4})^{3/4} + C\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2\[\int\limits_{} \frac{x^3 }{20x^3 u^{1/4}}du\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2i mistyped 15; meant 20 :) the x^3s cancel; pull out the 1/20; and integrate the u^(1/4)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2\[\frac{1}{20} \int\limits_{} u^{1/4} du \rightarrow \frac{1}{20} \frac{u^{(1+ 4)/4}}{(1+4)/4}\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2u^(3/4)/(3/4) 4 * 4root(u^3)  20 * 3 4root(u^3)  = 4root[5x^41]/15 5 * 3

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2forgot to ^3 my (5x^4  1)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2\[\frac{\sqrt[4]{(5x^4 1)^3}}{15} +C\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You are great!!!!! So final answer: \[\frac{1}{20} (5x^{4} 1)^{3/4} + C?\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2that looks comparabley good :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2ack!!...1/15.....not 1/20

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Fantastic!! lol  I almost messed that one up! Thank you so much! I have another problem I might post a new question so I can award medals  lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2im sure we will be around :)
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